Invert actual bits of a number

Given a non-negative integer n. The problem is to invert the bits of n and print the number obtained after inverting the bits. Note that the actual binary representation of the number is being considered for inverting the bits, no leading 0’s are being considered.
Examples :

Input : 11
Output : 4
(11)10 = (1011)
After inverting the bits, we get:
(0100)2 = (4)10.

Input : 10
Output : 5
(10)10 = (1010)2.
After reversing the bits we get:
(0101)2 = (101)2
= (5)10.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Method 1 (Using bitwise operators)

Prerequisite : Toggling k-th bit of a number

C++

 // CPP program to invert actual bits // of a number. #include using namespace std;    void invertBits(int num) {     // calculating number of bits     // in the number     int x = log2(num) + 1;        // Inverting the bits one by one     for (int i = 0; i < x; i++)         num = (num ^ (1 << i));          cout << num; }    // Driver code int main() {     int num = 11;      invertBits(num);     return 0; }

Java

 // Java program to invert  // actual bits of a number. import java.io.*;    class GFG {     static void invertBits(int num)     {         // calculating number of          // bits in the number         int x = (int)(Math.log(num) /                        Math.log(2)) + 1;                // Inverting the          // bits one by one         for (int i = 0; i < x; i++)          num = (num ^ (1 << i));                 System.out.println(num);     }            // Driver code     public static void main (String[] args)      {         int num = 11;          invertBits(num);     }    }    // This code is contributed // by Anuj_67

Python3

 # Python3 program to invert actual  # bits of a number.  import math    def invertBits(num):         # calculating number of bits      # in the number      x = int(math.log2(num)) + 1        # Inverting the bits one by one      for i in range(x):          num = (num ^ (1 << i))             print(num)     # Driver Code num = 11 invertBits(num)     # This code is contributed  # by Rituraj Jain

C#

 // C# program to invert  // actual bits of a number. using System;    class GFG {     static void invertBits(int num)     {         // calculating number of          // bits in the number         int x = (int)(Math.Log(num) /                         Math.Log(2)) + 1;                // Inverting the          // bits one by one         for (int i = 0; i < x; i++)          num = (num ^ (1 << i));                 Console.WriteLine(num);     }            // Driver code     public static void Main ()      {         int num = 11;          invertBits(num);     }    }    // This code is contributed // by Anuj_67

PHP



Output:

4

Time complexity : O(log n)
Auxiliary space : O(1)

Method 2 (Using Bitset)
Here we use the flip() of bitset to invert the bits of the number, in order to avoid flipping the leading zeroes in the binary representation of number, we have calculated the number of bits in the binary representation, and flipped only the actual bits of number.We have used to_ulong() to convert bitset to number.

C++

 // CPP program to invert actual bits // of a number. #include using namespace std;    void invertBits(int num) {     // calculating number of bits     // in the number     int x = log2(num) + 1;        // Considering number to be 32 bit integer;     bitset<32> b(num);        // reversing the bits one by one     for (int i = 0; i < x; i++)          b.flip(i);        // converting bitset to number     cout << b.to_ulong();  }    // Driver code int main() {     int num = 11;      invertBits(num);     return 0; }

Java

 // Java program to invert actual  // bits of a number. class GFG  {     static void invertBits(int num)     {         // calculating number of          // bits in the number         int x = (int)(Math.log(num) /                        Math.log(2)) + 1;                // Inverting the bits         // one by one                    for (int i = 0; i < x; i++)          num = (num ^ (1 << i));                 System.out.print(num);     }            // Driver code     public static void main(String[] args)     {         int num = 11;          invertBits(num);     } }    // This code is contributed by Mukul Singh

Python 3

 # Python 3 program to invert actual  # bits of a number. import math    def invertBits(num):            # calculating number of      # bits in the number     x = int(math.log(num, 2.0) + 1);        # Inverting the bits     # one by one     for i in range(0, x):          num = (num ^ (1 << i));         print(num);    # Driver code num = 11;  invertBits(num);    # This code is contributed  # by Akanksha Rai

C#

 // C# program to invert actual  // bits of a number. using System;    class GFG  {     static void invertBits(int num)     {         // calculating number of          // bits in the number         int x = (int)Math.Log(num, 2) + 1;                // Inverting the bits         // one by one         for (int i = 0; i < x; i++)          num = (num ^ (1 << i));                 Console.Write(num);     }            // Driver code     static void Main()     {         int num = 11;          invertBits(num);     }    }    // This code is contributed by Anuj_67

PHP



Output :

4

Time complexity: O(log n)
Auxiliary space: O(1)

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